Question

The value of $\int{\dfrac{\cos 2x}{\cos x}dx}$ is equal to:
(A). $2\sin -x+\log \left( \sec x-\tan x \right)+C$
(B). $2\sin -x\log \left( \sec x-\tan x \right)+C$
(C). $2\sin x+\log \left( \sec x+\tan x \right)+C$
(D). $2\sin x-\log \left( \sec x+\tan x \right)+C$

Answer 1

Hint: From the trigonometric identities, we know that $\cos 2x=2{{\cos }^{2}}x-1$. Substitute this value of $\cos 2x$ in the place of the numerator of the given integral and then we left with $2\cos x-\sec x$ inside the integral then integrate it with respect to x.

Complete step-by-step solution -
The integral given in the question is:
$\int{\dfrac{\cos 2x}{\cos x}dx}$
We can write the numerator of the above integral as $2{{\cos }^{2}}x-1$ because we know from the trigonometric identities that $\cos 2x=2{{\cos }^{2}}x-1$.
 $\int{\dfrac{\left( 2{{\cos }^{2}}x-1 \right)}{\cos x}dx}$
Simplifying the above integral as follows:
$\int{\left( 2\cos x-\dfrac{1}{\cos x} \right)dx}$
We know that reciprocal of $\cos x$ is $\sec x$ so substituting $\sec x$ in place of $\dfrac{1}{\cos x}$ in the above integral.$\int{\left( 2\cos x-\sec x \right)dx}$
Now separately integrating the trigonometric functions we get,
$\int{2\cos xdx-\int{\sec xdx}}$
We know that integration of $\cos x$ is $\sin x$ and integration of $\sec x$ is $\ln \left( \sec x+\tan x \right)$ so using these integral values in the above integral we get,
$2\sin x-\ln \left( \sec x+\tan x \right)+C$
From the above integration, the value of integral given in the question is found to be: $2\sin x-\ln \left( \sec x+\tan x \right)+C$
Hence, the correct option is (d).

Note: You might want to know how the integration of $\sec x$ is $\ln \left( \sec x+\tan x \right)$. In the below, we are going to see how this integral value came.
We know that the derivative of $\sec x+\tan x$is:
$\begin{align}
  & \dfrac{d\left( \sec x+\tan x \right)}{dx}={{\sec }^{2}}x+\sec x\tan x \\
 & \Rightarrow \dfrac{d\left( \sec x+\tan x \right)}{dx}=\sec x\left( \sec x+\tan x \right) \\
\end{align}$
As you see from the above that $\sec x+\tan x$ is common on both the sides of the equation so we will assume $\sec x+\tan x$ as “u”.
$\dfrac{du}{dx}=\sec x\left( u \right)$
Writing terms containing x on the one side of the equation and terms containing “u” on the other side of the equation we get,
$\dfrac{du}{u}=\sec xdx$
Integrating on the both the sides we get,
$\int{\dfrac{du}{u}}=\int{\sec xdx}$
$\ln u+C=\int{\sec xdx}$
Substituting the value of “u” as $\sec x+\tan x$ in the above equation we get,
$\ln \left( \sec x+\tan x \right)+C=\int{\sec xdx}$
Hence, we have derived the integration of $\sec x$.